SOLUTION: Beth is going to be in the big race next week. She will have to run and bike a total of 85 miles. She can run 6 mph and bike at 22 mph. During the race she will bike for 2 hours mo

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Question 1124196: Beth is going to be in the big race next week. She will have to run and bike a total of 85 miles. She can run 6 mph and bike at 22 mph. During the race she will bike for 2 hours more than she runs. Find the time she will spend running and the time she will spend biking.
Rate =X; formula used was d=rt
run 85=x(6mph) bike 85=(2+x)(22 mph)
I went wrong somewhere.
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let "x" be  the time she will spend biking (in hours).

Then the time running is (x+2) hours.



The distance running is  6*(x+2)  (speed times time).

The distance biking  is  22*x     (speed times time).



Your equation is


    Distanse + distance = total distance,   or


    6*(x+2)  + 22x      = 85  miles.


Simplify and solve for x.


Then evaluate (x+2).


Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Beth is going to be in the big race next week. She will have to run and bike a total of 85 miles. She can run 6 mph and bike at 22 mph. During the race she will bike for 2 hours more than she runs. Find the time she will spend running and the time she will spend biking.
Rate =X; formula used was d=rt
run 85=x(6mph) bike 85=(2+x)(22 mph)
I went wrong somewhere.
Apparently, you have the time spent running as x (Don’t enter x and X because they’re NOT the same, and x is NOT rate, but time, instead), and the time spent biking, 2 + x

These are correct!

Also, you have the distance running and biking as: 6x, and 22(2 + x), respectively.

These are correct also!

Now, remember the TOTAL DISTANCE traveled is 85 miles, so based on the TRAVEL FORMULA: d = rt, we get: 6x + 22(2 + x) = 85

Now solve this for x, the time spent running.

Add 2 to x to get the time spent biking.